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Question Number 119517 by bemath last updated on 25/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }[\sin (x+\frac{1}{x})-\sin x]=?$$

Answered by Dwaipayan Shikari last updated on 25/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }(sin(x+\frac{1}{x})-sinx)$$$$=\underset{x\to \mathrm{\infty }}{\lim }\left(2cos(x+\frac{1}{2x})sin\left(\frac{1}{x}\right)\right)=0$$$$-1⩽cos(x+\frac{1}{2x})⩽1$$$$\underset{x\to \mathrm{\infty }}{\lim }sin\frac{1}{x}\to 0$$

Answered by benjo_mathlover last updated on 25/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }[\sin x\cos \frac{1}{x}+\cos x\sin \frac{1}{x}-\sin x]=$$$$\underset{x\to \mathrm{\infty }}{\lim }[\sin x(\cos \frac{1}{x}-1)+\cos x\sin \frac{1}{x}]$$$$[note\underset{x\to \mathrm{\infty }}{\lim }\cos \frac{1}{x}-1=0\wedge \underset{x\to \mathrm{\infty }}{\lim }\cos x\sin \frac{1}{x}=0$$$$because\cos xoscilatesbetween-1and1]$$$$thus\underset{x\to \mathrm{\infty }}{\lim }[\sin x(\cos \frac{1}{x}-1)+\cos x\sin \frac{1}{x}]=0$$

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